The theory presented to me didn't make me understand how to do this kind of problem. I need help. Thanks you!
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Notice that $\frac{1}{2\cdot 3} = \frac{1}{2} - \frac{1}{3}$. – Winther Nov 12 '19 at 18:53
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Also: https://math.stackexchange.com/q/255306/42969, https://math.stackexchange.com/a/1652789/42969. – Martin R Nov 12 '19 at 18:54
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It's a telescoping sum:
$$\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{n(n+1)}$$
$$=\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)$$
$$=1-\frac1{n+1}=\dfrac n{n+1}$$
J. W. Tanner
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